This question already has an answer here:
I came across this problem while solving another one. I will show how far I could get on my own:
Suppose that $\sqrt 3 \in Q(\sqrt2)$. Since $Q(\sqrt3)$ is the smallest field containing both $Q$ and $\sqrt3$, thus $Q(\sqrt3)\subset Q(\sqrt2)$.
Now $[Q(\sqrt2):Q] =4 $ and $[Q(\sqrt3):Q]=2$ (I've already proven both), so$[Q(\sqrt2):Q(\sqrt 3)] =2$. Therefore by the definition of degree extension, there exist $p(x)\in Q(\sqrt 3)[X]$ of degree 2 realizing $\sqrt2$, that is, exists: $a,b,c\in Q(\sqrt 3)$ such that:
$$a(\sqrt2)^2+b \sqrt2+c =0.$$
How to proceed now? I should use the fact that those extensions lie in $\mathbb R$ to get a contradiction? I only know that $a\neq 0$, but what about the other coefficients?
EDIT: I could find an argument below, check the answers. Check it out if you agree! The argument is just a continuation of the above one.